- +1-530-264-8006
- info@tutorsglobe.com

18,76,764

Questions

Asked

21,311

Experts

9,67,568

Questions

Answered

Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!

Submit Assignment2015 © Tutors Globe. All rights reserved.

## Basic Thermodynamic Concepts, Physics tutorial

First Law of Thermodynamics:For a closed system that undergoes no changes in kinetic or potential energy, the differential form of the first law of thermodynamics states that

δQ = δU + δW

Here δQ is the change in the heat content of the system, δU is the change in the internal energy of the system, and δW is the element of work done by the system. The internal energy of the system δU is given by

δU = cvδT

Here cv is the specific heat capacity of the gas at constant volume, and δT is the change in the temperature of the gas. In general, the specific heat capacity at constant volume cv is given by

cv = (∂U/∂T)v

Here T is the temperature of the gas.

The element of work δW done by the system is given by

δW = pδV

Here p is the pressure of the gas, and δV is the change in the volume of the same gas . For an ideal gas, the equation of state is where n is the number of moles, R is the universal gas constant and T is the absolute temperature of the gas. The parameter R is numerically equal to 8.314 1 1J mol K.

In practice, a gas may expand (or be compressed) in different ways. The gas may also be heated (or be cooled) in different ways. In what follows, you shall be introduced to two different processes involving expansion, compression, heating and cooling of a gas. The two processes are the isothermal process and the adiabatic process.

Isothermal Process:An isothermal process is one which takes place at constant temperature. For one mole of an ideal gas, the equation of state may be written as

pV = RT

Here symbols have their usual meanings. In an isothermal process, the temperature T is constant. Thus, the equation of state of an ideal gas undergoing isothermal process is

pV = k

Here k is a constant. Hence, you may write

p = k/V

This implies that the p-V curve must be a rectangular hyperbola. In an isothermal expansion, work is done by the system on the surroundings.

The element of work δW done by an ideal gas undergoing isothermal expansion is

δW = pδV

Therefore, the work done in expanding from V

_{1}to V_{2}isW = ∫dW

W =

_{v1}∫^{v2}pdVW =

_{v1}∫^{v2}(RT/V)dVW = RT

_{v1}∫^{v2}dV/VW = RT[logeV]

_{v1}^{v2}W = RTlog

_{e}(V_{2}/V_{1})Work W done by gas is equal to quantity of heat Q required to do work. Therefore, heat Q needed is also given by

Q = RTlog

_{e}(V_{2}/V_{1})Now for isothermal process,

pV = k

Here k is a constant.

p

_{1}V_{1 }= p_{2}V_{2}V

_{2}/V_{1}= p_{1}/p_{2}Therefore, work done, in isothermal process, may also be written as

W = Q = RTlog

_{e}(p_{1}/p_{2})This quantity of heat Q should be supplied to system that is doing work on surroundings to keep temperature constant.

Otherwise, temperature of system falls, and procedures stops to be isothermal. In case of isothermal compression, work is done on system by the surroundings. In this case, heat Q should be removed from system to keep its temperature constant. Otherwise, temperature rises, and procedures stops to be isothermal.

Perfect isothermality is a highly idealized concept. It is not possible to realize it in practice. For example, heat Q can't flow through walls of vessel unless there is temperature difference across walls. Thus, temperature of gas is bound to fall little in isothermal expansion. Also, temperature of gas is bound to raise little in an isothermal compression. Though, if procedure is slow, and gas is held in thin, well-conducting vessel surrounded by the constant-temperature bath, procedure becomes around isothermal.

Adiabatic Process:Equation of Reversible Adiabatic Change:The adiabatic procedure is one which occurs at constant heat. In adiabatic process, temperature may change but no heat enters or leaves system. Consider the adiabatic process in which unit mass of ideal gas expands from initial volume V

_{1}to final volume V_{2}. First law of thermodynamics signifies that heat dQ supplied to system is given bydQ = dU + dW

Where dU is increase in internal energy of the system and dW is external work done by system.

For a reversible adiabatic change dQ= 0.

Thus equation reduces to

c

_{v}dT + pdV = 0dT = -pdV/c

_{v}Recall that the equation of state for a unit mass of an ideal gas is

pV = RT

Differentiating Equation you obtain

pdV + Vdp = RdT

dT = (pdV + Vdp)/R

c

_{p}and c_{v}are related to gas constant R in given wayR = c

_{p}- c_{v}Using equations:

c

_{p}pdV + c_{v}Vdp + (c_{p}- c_{v})pdV = 0Vdp + (c

_{p}/c_{v})pdV = 0If set g = c

_{p}/c_{v}we getVdp + gpdV = 0

Dividing each term by pV, you get

dp/p + g(dV/V) = 0

Integrating, find that:

O'(dp/p) + gO'(dV/V) = constant or log

_{e}P + glog_{e}V = aWhere a is a constant

Applying elementary laws of logarithm, we get:

pV

^{g}= e^{a}Thus, you may write:

pV

^{g}constant,As a is a constant.

Equation is the equation of the reversible adiabatic process.

Temperature Change in an Adiabatic Process:Recall that the general gas equation is

pV = RT

So that

p = RT/V

pV

^{g}= RTV^{g}V^{-1}Or

pV

^{g}= RTV^{g-1}But the equation of an adiabatic process is

pV

^{g}= constant,RTV

^{g-1}= constantThis is the equation for adiabatic temperature change.

Work Done in Adiabatic Process:The work W done by a unit mass of gas undergoing adiabatic expansion is given by

W =

_{V1}O'^{V2}pdVAs process is adiabatic, it follows that

pV

^{g}= bHere b is constant given by b = e

^{a}. Thereforep = bV

^{-g}Therefore, work W becomes

W =

_{v1}O'^{v2}bV^{-g}dVW = (b/1-g)(V

_{2}^{1-g }- V_{1}^{1-g})After solving

pV = bV

^{-g+1}By expanding

W = (1/1-g)(bV

_{2}^{1-g }- bV_{1}^{1-g})So that

W = (1/1-g)(p

_{2}V_{2 }- p_{1}V_{1})Since, pV = bV

^{-g+1}This equation represents work W done by system when unit mass of ideal gas expands adiabatically from initial volume V

_{1 }to final volume V_{2}.Enthalpy and Internal Energy:The first law of thermodynamics may be defined as follows:

dQ= dU + pdV

If pressure p is constant then there is actually no difference between pdV and d(pV).

For example,

d(pV) = pdV + Vdp and if pressure p is constant, then dp = o so that

d(pV) = pdV + 0

By stating new thermodynamic property H such that.

H = U + pV

This new thermodynamic property H is known as enthalpy of system. It follows, thus, that

dH = c

_{p}dTNow compare Equation with expression for change in internal energy dU of system, that is,

dU = c

_{v}dTc

_{v}= ∂U/∂TThis equation illustrates that specific heat capacity at constant volume c

_{v }may be written as:c

_{v}= ∂U/∂TThis equation illustrates that specific heat capacity at constant pressure c

_{p}may be written asc

_{p}= ∂H/∂TSpecific Heats for Ideal Gases:Useful relation exists between principal specific heats and universal gas constant R. This relation may be defined as follows:

R = c

_{p}- c_{v}Change in internal energy of system is provided by dU = cvdT

Also, change in enthalpy of system is

dH = c

_{p}dTBy subtracting equation we get

dH - dU = (c

_{p}- c_{v})dTAs

pV = RT

dH - dU = RdT

After solving we get:

R = c

_{p }- c_{v}Equation given above illustrates that difference of principal specific heat capacities is numerically equal to gas constant.

Entropy and Second Law of Thermodynamics:Second law of thermodynamics may be defined as:

a) It is impossible to build device that operates in cycle and whose sole effect is to transfer heat from cooler body to hotter body. (This is Clausius statement of second law of thermodynamics).

b) It is not possible to build the device that operates in cycle and generates no other effect than production of work and exchange of heat with single reservoir. (This is Kelvin-Plank statement of second law of thermodynamics).

Clausius statement of second law of thermodynamics illustrates heat flows from a high-temperature region to low temperature region in absence of other effects. Therefore, when the hot body is brought in contact with cold body, hot body becomes less hot and cold body becomes less cold. It is impossible for cold body (in contact with a hot body) to be become more cold unless extra effects are brought to bear on system.

Quantity dQ/T is exact differential that may be denoted by ds. In this case, ds = dQ/T where s is another new property of system called entropy. It is significant to note that entropy is so stated only for reversible processes. For such reversible procedure a change in entropy may be with written as

V

_{s}= s_{2}- s_{1 }=_{a1}∫^{a2}dQ/THere a

_{1}and a_{2}are the suitable lower and upper limits of integration respectively.The equality holds for reversible process and inequality holds for irreversible process.

The reversible procedure is idealized state path. Procedure is said to be reversible if initial state of system can be restored following exactly same state path in reverse direction with no observable effects on system and its surroundings.

Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.

Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.

Free to know our price and packages for online physics tutoring. Chat with us or submit request at info@tutorsglobe.com